Dijkstra’s Algorithm Keep Going!

Pre-Computing Shortest Paths How many paths to pre-compute? Recall: –Using single-source to single-dest find_path: Need any delivery location to any other travel time: –N * (N-1)  2450 calls for N = 50 Plus any depot to any delivery location –M * N  500 calls for N = 50, M = 10 Plus any delivery location to any depot –N * M  500 calls –Total: 3450 calls to your find_path

Pre-Computing Travel Time Paths Using single-source to all destinations –Need any delivery location to any other N calls  50 –Plus any depot to any delivery location M calls  10 –Plus any delivery location to any depot 0 calls –Total: 60 calls Is this the minimum? –No, with small change can achieve: 51 calls Get from earlier call

Is This Fast Enough? Recall: –Dijkstra’s algorithm can search whole graph –Especially with multiple destinations –O(N) items to put in wavefront –Using heap / priority_queue: O (log N) to add / remove 1 item from wavefront Total: –N log N –Can execute in well under a second –OK!

Escaping Local Minima Revisited

Say We’re In This State Local perturbation to improve? deliveryOrder = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Swap Order of Two Deliveries? deliveryOrder = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} deliveryOrder = {0, 1, 3, 2, 4, 5, 6, 7, 8, 9} No swap of two deliveries can improve! Stuck in a local minimum

2-Opt? Path cut into 3 pieces deliveryOrder = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

2-Opt? Reconnected: worse! deliveryOrder = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}deliveryOrder = {0, 1, 2, 6, 5, 4, 3, 7, 8, 9}

2-Opt? Reconnected differently: now better! deliveryOrder = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}deliveryOrder = {0, 1, 2, 7, 8, 9, 6, 5, 4, 3}

Perturbations & Local Minima Explore lots of local perturbations –Compute travel time for each –To see what’s better Escape local minima with more powerful perturbations And/or use high climbing Powerful unifying technique –Simulated annealing –Lots of high climbing early –Little later (metal has cooled)

How Do I Finish by the Time Limit? #include #define TIME_LIMIT 30 // m4: 30 second time limit int main ( ) { clock_t startTime = clock (); // Clock “ticks” do { myOptimizer (); clock_t currentTime = clock (); float timeSecs = ((float) (currentTime – startTime)) / CLOCKS_PER_SEC; // Keep optimizing until within 10% of time limit } while (timeSecs < 0.9 * TIME_LIMIT);... }

Algorithm Challenge

Algorithms: Challenge Question Frank likes what he calls “cool” numbers For cool numbers, there are integers x and y such that –Cool number = x 2 = y 3 –For example, 1 is cool (= 1 2 = 1 3 ) and 64 is cool (= 8 2 = 4 3 ) –25 is not cool (= 5 2, but no integer cubed = 25) 1.Write a program to print all cool numbers between 1 and N 2.Calculate the computational complexity of your program 3.Mail me program & complexity: first 5 of lowest complexity  chocolate bar in class Fri. Source: ACM Programming Competition

Multithreading Why & How

Intel 8086 First PC microprocessor ,000 transistors 5 MHz ~10 clocks / instruction ~500,000 instructions / s

Intel Core i billion transistors 3.5 GHz ~15 clocks / instruction, but ~30 instructions in flight at once  Average about 2 instructions completed / clock Can execute ~7 billion instructions / s

1978 to ,000x more transistors ~14,000x more instructions / s The future: –Still getting 2X the transistors every 2 years –But transistors not getting much faster Clock speed saturating –~30 instructions in flight Complexity & power to go beyond this climbs rapidly Slow growth in instructions / cycle –Impact: CPU speed not increasing as rapidly Using multiple processors (cores) now important Multithreading: one program using multiple cores at once

A Single-Threaded Program Instructions (code) Memory Global Variables Heap Variables (new) Stack (local variables)... Program Counter Stack Pointer CPU / Core

A Multi-Threaded Program Instructions (code) Memory Global Variables Heap Variables (new) Stack1 (local variables)... Program Counter Stack Pointer Core1 Program Counter Stack Pointer Core2 Stack2 (local variables) thread 1 thread 2 Shared by all threads Each thread gets own local variables

Thread Basics Each thread has own program counter –Can be executing a different function –Is (almost always) executing a different instruction from other threads Each thread has own stack –Has its own copy of local variables (all different) Each thread sees same global variables Dynamically allocated memory –Shared by all threads –Any thread with a pointer to it can access

Implications Threads can communicate through memory –Global variables –Dynamically allocated memory –Fast communication! Must be careful threads don’t conflict in reads/write to same memory –What if two threads update the same global variable at the same time? –Not clear which update wins! Can have more threads than CPUs –Time share the CPUs